## Calculus (3rd Edition)

$$y=e(x+2).$$
We have $$y'=e^{x+2}\Longrightarrow y'(-1)=e.$$ Then the slope of the tangent line at $x=-1$ is $m= e$. Hence the equation of the tangent line is given by $$y=mx+c=ex+c.$$ Since $y(-1)=e$, then $c=2e$, hence the tangent line at $x=-1$ is given by $$y=ex+2e=e(x+2).$$